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Step 1: Understanding the System
We have two metallic balls with charges $q_1$ and $q_2$, and radii $R_1$ and $R_2$. The total charge $Q$ of the system is given by:
$ Q = q_1 + q_2 $
Assume that the potential of each ball is equal because they are in equilibrium due to symmetry, hence we denote the potential of each ball as $V$.

Step 2: Potential of a Metallic Sphere
The potential $V$ of a charged metallic sphere is given by:
$ V = \frac{k imes q}{R} $
where $k$ is Coulomb's constant. Therefore, for the two balls, we have:
$ V_1 = \frac{k imes q_1}{R_1} $
$ V_2 = \frac{k imes q_2}{R_2} $
Since they are equal at equilibrium, we have:
$ \frac{k imes q_1}{R_1} = \frac{k imes q_2}{R_2} $
Thus, removing $k$ from both sides, we find:
$ \frac{q_1}{q_2} = \frac{R_1}{R_2} $

Step 3: Minimizing the Energy
The electrostatic potential energy $U$ of the system consisting of two charges is given by:
$ U = \frac{k \times q_1 \times q_2}{d} $
where $d$ is the distance between the centers of the balls. However, to minimize energy concerning the ratio of charges, we find that the best configuration occurs when the ratio of charges equals the ratio of radii.

Step 4: Potential Difference
To find the potential difference $V_{12}$ between the two balls, we use the potentials previously found:
$ V_{12} = V_1 - V_2 $
This gives us:
$ V_{12} = \frac{k \times q_1}{R_1} - \frac{k \times q_2}{R_2} $
Rewriting $q_2$ in terms of $q_1$ using the ratio found, we get:
$ V_{12} = k \left( \frac{q_1}{R_1} - \frac{q_2}{R_2} \right) = k \left( \frac{q_1}{R_1} - \frac{q_1 \frac{R_1}{R_2}}{R_2} \right)$
Thus, simplifying further, we find:
$ V_{12} = k \times q_1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $
By substituting $q_2 = Q - q_1$, we can explore various configurations to define the energy minimum which is consistent with the initial ratio of charges.

Therefore, the ratio $\frac{q_1}{q_2} = \frac{R_1}{R_2}$ is maintained for energy minimization and confirms that energy is more optimally distributed at electric equilibrium.